Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{-z^2 + 9z}{z + 3} \div \dfrac{z^2 - 3z - 54}{-9z - 27} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{-z^2 + 9z}{z + 3} \times \dfrac{-9z - 27}{z^2 - 3z - 54} $ First factor the quadratic. $a = \dfrac{-z^2 + 9z}{z + 3} \times \dfrac{-9z - 27}{(z - 9)(z + 6)} $ Then factor out any other terms. $a = \dfrac{-z(z - 9)}{z + 3} \times \dfrac{-9(z + 3)}{(z - 9)(z + 6)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ -z(z - 9) \times -9(z + 3) } { (z + 3) \times (z - 9)(z + 6) } $ $a = \dfrac{ 9z(z - 9)(z + 3)}{ (z + 3)(z - 9)(z + 6)} $ Notice that $(z + 3)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 9z\cancel{(z - 9)}(z + 3)}{ (z + 3)\cancel{(z - 9)}(z + 6)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $a = \dfrac{ 9z\cancel{(z - 9)}\cancel{(z + 3)}}{ \cancel{(z + 3)}\cancel{(z - 9)}(z + 6)} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $a = \dfrac{9z}{z + 6} ; \space z \neq 9 ; \space z \neq -3 $